A particle of mass 'm' executes simple harmonic motion with amplitude 'a' and frequency v. The average kinetic energy during its motion from the position of equilibrium to the end is :

Options

A. `(1)/(4)ma^(2)v^(2)`
B. `4pi^(2)ma^(2)v^(2)`
C. `2pi^(2)ma^(2)v^(2)`
D. `pi ma^(2)v^(2)`.

Answer

D

Explanation

The average value of kinetic energy during motion from mean to extreme position `=(1)/(2)((1)/(2)m omega^(2)a^(2)+0)`
`=(1)/(4)m(2pi v)^(2).a^(2)=pi ma^(2)v^(2)`
So correct choice is (d).